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php - 使用 MySQL 创建触发器

转载 作者:行者123 更新时间:2023-11-29 23:16:22 26 4
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两个表:用户和消息

 CREATE TABLE IF NOT EXISTS `users` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`username` varchar(10) NOT NULL,
`password` varchar(10) NOT NULL,
`email` varchar(100) NOT NULL,
`verifystring` varchar(20) NOT NULL,
`active` tinyint(4) NOT NULL,
`usertype` tinyint(4) NOT NULL,
`img` varchar(200) DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=8 ;

CREATE TABLE IF NOT EXISTS `messages` (
`id` tinyint(4) NOT NULL AUTO_INCREMENT,
`date` datetime NOT NULL,
`user_id` int(11) NOT NULL,
`topic_id` int(11) NOT NULL,
`subject` varchar(100) NOT NULL,
`body` text NOT NULL,
`img` varchar(200) DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `topic_id` (`topic_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1 AUTO_INCREMENT=51 ;

触发:

CREATE TRIGGER `updateimg` AFTER UPDATE ON `users`
FOR EACH ROW begin
update messages set img = new.img where messages.user_id = users.id;
end

img 字段保存用户上传的图像的名称。更新时如何将值 img 从表 users 复制到表 messages ?

最佳答案

应该这样做:

CREATE TRIGGER update AFTER UPDATE ON table1
for each ROW
begin
update table2 set y = new.y where table2.user_id = new.id;
end

关于php - 使用 MySQL 创建触发器,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/27761150/

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